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3 years ago

If a triangle is a right triangle, then the other two angles must be acute. true or false

1 answer:

5 0

That depends on which angle you picked first, because that determines
which angles "the other two" are.

If you picked the right angle (90°) first, before you asked the question,
then the other two are acute angles, and they're also complementary

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Bambi the young dear was distracted Buy butterfly and jumped into the road in front of the two vehicles as shown in the diagram

bagirrra123 [75]

Speed of car A is given as

$v_a = 70 mph$

now we need to convert it into SI units

1 miles = 1609 m

1 hour = 3600 s

now we have

$v_a = 70 *\frac{1609}{3600} = 31.3 m/s$

now its distance from Bambi is given as

$d_a = 350 m$

time taken by it to hit the Bambi

$t = \frac{d}{v}$

$t = \frac{350}{31.3}$

$t = 11.2 s$

Now other car is moving at speed 50 mph

so its speed in SI unit will be

$v_b = 50* \frac{1609}{3600}$

$v_b = 22.35 m/s$

now its distance from Bambi is given as

$d_b = 590 feet$

as we know that 1 feet = 0.3048 m

$d_b = 590*0.3048 = 179.83 m$

now the time to hit the other car is

$t_2 = \frac{179.83}{22.35}$

$t_b = 8.05 s$

So Car B will hit the Bambi first

7 0

3 years ago

How does vegetation in the soil affect the rate of erosion

avanturin [10]

Answer:

Plants slow down water as it flows over the land and this allows much of the rain to soak into the ground. Plant roots hold the soil in position and prevent it from being blown or washed away. Plants break the impact of a raindrop before it hits the soil, reducing the soil's ability to erode.

Explanation:

4 0

3 years ago

two ships leave a port at the same time. The first ship sails on a bearing of 40 degrees at 18 knots and the second at a bearing

Yuliya22 [10]

Answer:

The distance between the ships is 87.84 km.

Explanation:

Given that,

Angle of first ship= 40°

Speed of first ship = 18 knots

Angle of second ship= 130°

Speed of second ship = 26 knots

We need to calculate the resultant velocity

Using cosine rule

$v=\sqrt{v_{1}^2+v_{2}^2-2v_{1}v_{2}\cos\theta}$

Put the value into the formula

$v=\sqrt{18^2+26^2-2\times18\times26\times\cos90}$

$v=\sqrt{18^2+26^2}$

$v=\sqrt{324+676}$

$v=10\sqrt{10}$

We need to calculate the distance between the ships

$d =v\times t$

Put the value into the formula

$d=10\sqrt{10}\times1.5\times1.852$

$d=87.84\ km$

Hence, The distance between the ships is 87.84 km.

7 0

3 years ago

2. a) Force affects the body and results in different changes. The grid given below

postnew [5]

Answer:

Please find the answer in the explanation

Explanation:

Given that Force affects the body and results in different changes.

The grid that has the words related to the effects of force are :

a) Force could be a push or a pull.

c) Force has magnitude as well as direction.

e) Force acting on an object may cause a change in its state of motion or a change in its shape.

f) A force can act on an object with or without being in contact with it.

8 0

3 years ago

three girls were pushing the same car with a net force of 450 N [N48°E]. Two of the girls were pushing with forces of 310 N [N25

ElenaW [278]

The net force is the vector

∑ F = (450 N) (cos(42°) i + sin(42°) j)

and two of the forces provided by the girls are

F₁ = (310 N) (cos(115°) i + sin(115°) j)

F₂ = (250 N) (cos(285°) i + sin(285°) j)

Then the force provided by the third girl is the vector

F₃ = ∑ F - F₁ - F₂

F₃ = ((450 N) cos(42°) - (310 N) cos(115°) - (250 N) cos(285°)) i

… … … + ((450 N) sin(42°) - (310 N) sin(115°) - (250 N) sin(285°)) j

F₃ ≈ (400.722 N) i + (261.635 N) j

So, the third girl provided a force of magnitude

||F₃|| = √((400.722 N)² + (261.635 N)²) ≈ 478.572 N ≈ 480 N

pointing in a direction

arctan((261.635 N)/(400.722 N)) ≈ 33.1409° ≈ 33°

relative to East which refers to 0°; that is, 33° N of E or E33°N. Since the other forces are given relative to North or South, we can write this direction as N57°E.

So, the third girl pushed with force 480 N [N57°E].

5 0

2 years ago